Question

A ball is thrown from a point at a
distance 40 m from a wall of height
15 m. It just clears the wall and then
attains maximum height. Find the
maximum height if the angle of
projection is 45°​

Answer 1

Explanation:

y = x tan theta - x tan ^2 theta /R

Put theta = 45

→ 15 = 40 - 40^2/R

→ -25 = -40^2 /R

→ R = 64

We know that ,

R / H = 4/ tan theta

→ 64/H = 4/ 1

→ H = 16

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